Hydrostatic Equilibrium and Atmospheric Pressure

Recommended resources

The freely available resources this lesson will direct you to.

Hello! Welcome to the first lesson of our second module, "Radiative Transfer and Atmospheric Structure."

In Module 1, we established the foundational physics of Earth's energy balance, treating the atmosphere as a simple, uniform slab. Now, we will begin to add layers of complexity and realism. To understand how radiation travels through the atmosphere, we first need to understand the atmosphere's physical structure—specifically, how its pressure and density change with height.

Today's lesson addresses the first learning outcome of this module: to derive the hydrostatic equilibrium equation and use it to explain the vertical pressure structure of the atmosphere. We will explore the fundamental balance of forces that supports the atmosphere against gravity and derive the mathematical relationship that governs its vertical structure.

1. The Balance of Forces in the Atmosphere

Imagine the entire atmosphere. Gravity is constantly pulling every air molecule towards the Earth's surface. Why doesn't it all collapse into an incredibly dense, thin layer? The answer lies in pressure. The atmosphere has weight, and this weight creates pressure. Crucially, this pressure is higher at lower altitudes and lower at higher altitudes. This difference in pressure creates an upward force that counteracts gravity.

This state of balance is called hydrostatic equilibrium.

To begin, let's build an intuition for atmospheric pressure and the forces it generates.

Hydrostatic Equilibrium and Exponential Atmospheres

This brief reading from an astronomy course at the University of Maryland provides an excellent conceptual introduction to atmospheric pressure and why pressure gradients, not pressure itself, are what generate forces.

Please read the introductory section (points 1-5) on the first page. Focus on the distinction between pressure and a pressure gradient.

The key takeaway is that a net force arises only from a difference in pressure. In the vertical direction, the pressure at the bottom of any given parcel of air is slightly higher than the pressure at its top, resulting in a net upward push.

This image provides a great visual summary of the concept.

This diagram illustrates hydrostatic equilibrium. The upward pressure gradient force, caused by pressure being higher at lower altitudes (P1 > P2), balances the downward force of gravity on a parcel of air. The equation shown, Δp/Δz = ρg, is a simplified form of the hydrostatic equation we are about to derive.

2. Deriving the Hydrostatic Equilibrium Equation

Now, let's formalize this concept by deriving the governing equation. Your background in physics and calculus will be very useful here. We will consider the forces acting on an infinitesimally thin, static slab of air.

The two forces acting on this slab in the vertical direction are:

Gravity: Pulling the slab downward.
Pressure: Pushing on the top and bottom surfaces of the slab.

Let's define our slab of air:

It has a horizontal area $A$ .
It has an infinitesimal thickness $d z$ .
It is located at an altitude $z$ .
The air within it has a uniform density $ρ$ .

The downward gravitational force ( $F_{g}$ ) is the mass of the slab multiplied by the acceleration due to gravity, $g$ .

Mass = Density × Volume = $ρ \times (A \cdot d z)$
So, $F_{g} = - (ρ A d z) g$ . We use a negative sign to indicate the downward direction.

The net pressure force ( $F_{p}$ ) is the difference between the upward force on the bottom surface and the downward force on the top surface.

Upward force on the bottom surface (at height $z$ ): $P (z) \cdot A$
Downward force on the top surface (at height $z + d z$ ): $P (z + d z) \cdot A$
Net force: $F_{p} = P (z) A - P (z + d z) A = - [P (z + d z) - P (z)] A$

For an infinitesimally thin slab, the term $P (z + d z) - P (z)$ is, by the definition of the derivative, equal to $d P$ . So, we can write the net pressure force as:
$F_{p} = - A \cdot d P$

In hydrostatic equilibrium, the net force is zero, so the upward pressure force must exactly balance the downward gravitational force:
$F_{p} + F_{g} = 0$
$- A \cdot d P - (ρ A d z) g = 0$

We can cancel the area $A$ from both terms and rearrange to find the hydrostatic equilibrium equation:
$\frac{d P}{d z} = - ρ g$

This elegant equation is fundamental to atmospheric science. It states that the rate of change of pressure with altitude ( $\frac{d P}{d z}$ ) is equal to the negative of the local air density times the acceleration due to gravity.

To see the derivation presented in a slightly different way and to reinforce your understanding, please review the following resource.

Hydrostatic Equilibrium and Exponential Atmospheres

This resource from the University of Maryland provides a concise, step-by-step derivation of the hydrostatic equilibrium equation using the same differential approach we just followed.

Read the section titled '1. Derivation of the Equation of Hydrostatic Equilibrium'. It walks through the same logic of balancing pressure and gravitational forces on a thin slab of atmosphere.

It's also useful to see this expressed in terms of finite differences, which is common in numerical modeling and practical measurements. The resource below uses this approach and includes a helpful diagram.

1.9: Hydrostatic Equilibrium

This excerpt from a practical meteorology textbook derives the equation using finite differences (ΔP/Δz) and explicitly discusses the relationship between the differential and finite-difference forms.

First, review the initial paragraphs and Figure 1.12 to see the forces on a slice of air. Then, read the derivation leading to equation 1.25b. Finally, read the 'HIGHER MATH' section at the end, which connects the finite-difference form (eq. 1.25b) to the differential form (eq. 1.25c) we derived.

3. Explaining the Vertical Pressure Structure

The hydrostatic equation $\frac{d P}{d z} = - ρ g$ directly explains the basic vertical pressure structure of the atmosphere.

Since density ( $ρ$ ) and gravity ( $g$ ) are always positive, the right-hand side is always negative.
This means the derivative $\frac{d P}{d z}$ is always negative, proving that pressure must decrease with increasing altitude.
Furthermore, it tells us that the rate of pressure decrease is proportional to the local air density. Near the surface, where air is densest, pressure falls off rapidly with height. At higher altitudes, where the air is much thinner, you must ascend much farther for the same pressure drop.

To move from this qualitative description to a quantitative model of the pressure profile, we need to solve the differential equation. As you know from your work with differential equations, this requires a relationship between the variables—in this case, between pressure $P$ and density $ρ$ . This relationship is called an equation of state.

For the atmosphere, the ideal gas law is an excellent approximation. In a form convenient for our purposes, it is:
$P = \frac{ρ k _{B} T}{m ˉ}$
where $k_{B}$ is the Boltzmann constant, $T$ is temperature, and $m ˉ$ is the mean molecular mass of air. Rearranging for density gives $ρ = P \frac{m ˉ}{k _{B} T}$ .

Substituting this into the hydrostatic equation:
$\frac{d P}{d z} = - (P \frac{m ˉ}{k _{B} T}) g$
$\frac{1}{P} d P = - (\frac{m ˉ g}{k _{B} T}) d z$

If we make the simplifying assumption that temperature $T$ and gravity $g$ are constant with altitude, this equation can be easily integrated to describe the full pressure profile.

Hydrostatic Equilibrium and Exponential Atmospheres

Let's now see how solving this differential equation leads to a specific functional form for the pressure profile. This reading walks through the integration, assuming an ideal gas and constant temperature.

Read section '2. Derivation of Pressure Profile' to see how the hydrostatic equation is solved to yield an exponential pressure decay. Also, read the following section that critically discusses the assumptions made (constant T, constant g, ideal gas), which should resonate with your background in statistical modeling.

As the reading shows, solving the equation under these assumptions yields:
$P (z) = P_{0} e^{- z / H}$
where $P_{0}$ is the pressure at a reference height $z = 0$ (e.g., sea level), and $H = \frac{k _{B} T}{m ˉ g}$ is a constant called the scale height.

This result is the quantitative explanation for the atmosphere's vertical pressure structure: pressure decreases approximately exponentially with altitude. The scale height, $H$ , represents the vertical distance over which pressure drops by a factor of $1/ e$ (about 37%). For Earth, the scale height is roughly 8.5 km.

Conclusion

In this lesson, we have built a physical and mathematical model for the atmosphere's vertical structure, moving beyond the simple "slab" model of Module 1.

Here are the key takeaways:

The atmosphere is held in hydrostatic equilibrium, where the upward-directed pressure gradient force perfectly balances the downward force of gravity.
This balance is mathematically expressed by the hydrostatic equation: $\frac{d P}{d z} = - ρ g$ .
This equation explains why atmospheric pressure decreases with altitude. By combining it with the ideal gas law and assuming a constant temperature, we find that pressure decreases exponentially with height, a profile characterized by the scale height $H$ .

This understanding of the vertical pressure and density structure is a critical prerequisite for the topics to come in this module.

Preview of the next lesson: We will formalize the exponential pressure profile as the barometric formula and explore the implications of the scale height. Following that, we will turn our attention to the atmosphere's temperature structure by deriving the dry adiabatic lapse rate.